$ D = \left[\begin{array}{rr}0 & 4 \\ 3 & -2\end{array}\right]$ $ A = \left[\begin{array}{rr}3 & -2 \\ 3 & 4\end{array}\right]$ What is $ D A$ ?
Because $ D$ has dimensions $(2\times2)$ and $ A$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ D A = \left[\begin{array}{rr}{0} & {4} \\ {3} & {-2}\end{array}\right] \left[\begin{array}{rr}{3} & \color{#DF0030}{-2} \\ {3} & \color{#DF0030}{4}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ A$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ A$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ A$ , and so on. Add the products together. $ \left[\begin{array}{rr}{0}\cdot{3}+{4}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{0}\cdot{3}+{4}\cdot{3} & ? \\ {3}\cdot{3}+{-2}\cdot{3} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{0}\cdot{3}+{4}\cdot{3} & {0}\cdot\color{#DF0030}{-2}+{4}\cdot\color{#DF0030}{4} \\ {3}\cdot{3}+{-2}\cdot{3} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{0}\cdot{3}+{4}\cdot{3} & {0}\cdot\color{#DF0030}{-2}+{4}\cdot\color{#DF0030}{4} \\ {3}\cdot{3}+{-2}\cdot{3} & {3}\cdot\color{#DF0030}{-2}+{-2}\cdot\color{#DF0030}{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}12 & 16 \\ 3 & -14\end{array}\right] $